Two capacitors, when connected in series, have an equivalent capacitance of 3 μF, and when they are connected in parallel their equivalent capacitance is 16 μF. Their values are respectively μF and μF.

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(4) 4, 12. Given: series value Cₛ = 3 μF, parallel value Cₚ = 16 μF. Idea: Cₚ = C₁ + C₂ and Cₛ = (C₁C₂)/(C₁ + C₂), so C₁C₂ = Cₛ Cₚ. So C₁ + C₂ = 16 and C₁C₂ = 3×16 = 48. C₁ and C₂ are the roots of t² - 16t + 48 = 0, i.e. (t-4)(t-12) = 0. Check: 4 + 12 = 16 and (4×12)/(16) = 3. So the capacitors are 4 μF and 12 μF.

Two capacitors, when connected in series, have an equivalent capacitance of 3 μF, and when they are connected in parallel their equivalent capacitance is 16 μF. Their values are respectively μF and μF.

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Two capacitors, when connected in series, have an equivalent capacitance of 3 μF, and when they are connected in parallel their equivalent capacitance is 16 μF. Their values are respectively ______ μF and ______ μF.

Step-by-step solution

Why the other options are wrong

Two capacitors, when connected in series, have an equivalent capacitance of 3 μF, and when they are connected in parallel their equivalent capacitance is 16 μF. Their values are respectively μF and μF.